Problem: Balance the following chemical equation: $ $ $\text{As}_4\text{S}_6 +$ $\text{O}_2 \rightarrow$ $\text{As}_4\text{O}_6 +$ $\text{SO}_2$
Solution: $\text{As}$ is already balanced. There are $6 \text{ S}$ on the left and only $1$ on the right, so multiply $\text{SO}_2$ by ${6}$ $ \text{As}_4\text{S}_6 + \text{O}_2 \rightarrow \text{As}_4\text{O}_6 + {6}\text{SO}_2 $ That gives us $18 \text{ O}$ on the right and only $2$ on the left, so multiply $\text{O}_2$ by ${9}$ . (Since oxygen is by itself on the left, it should be done at the end because you can give it a coefficient without affecting another element.) $ \text{As}_4\text{S}_6 + {9}\text{O}_2 \rightarrow \text{As}_4\text{O}_6 + 6\text{SO}_2 $ The balanced equation is: $ \text{As}_4\text{S}_6 + 9\text{O}_2 \rightarrow \text{As}_4\text{O}_6 + 6\text{SO}_2 $